Question: Solve for $x$ : $ 8|x + 10| - 1 = -4|x + 10| + 3 $
Solution: Add $ {4|x + 10|} $ to both sides: $ \begin{eqnarray} 8|x + 10| - 1 &=& -4|x + 10| + 3 \\ \\ { + 4|x + 10|} && { + 4|x + 10|} \\ \\ 12|x + 10| - 1 &=& 3 \end{eqnarray} $ Add ${1}$ to both sides: $ \begin{eqnarray} 12|x + 10| - 1 &=& 3 \\ \\ { + 1} &=& { + 1} \\ \\ 12|x + 10| &=& 4 \end{eqnarray} $ Divide both sides by ${12}$ $ \dfrac{12|x + 10|} {{12}} = \dfrac{4} {{12}} $ Simplify: $ |x + 10| = \dfrac{1}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 10 = -\dfrac{1}{3} $ or $ x + 10 = \dfrac{1}{3} $ Solve for the solution where $x + 10$ is negative: $ x + 10 = -\dfrac{1}{3} $ Subtract ${10}$ from both sides: $ \begin{eqnarray} x + 10 &=& -\dfrac{1}{3} \\ \\ {- 10} && {- 10} \\ \\ x &=& -\dfrac{1}{3} - 10 \end{eqnarray} $ Change the ${ - 10}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{1}{3} {- \dfrac{30}{3}} $ $ x = -\dfrac{31}{3} $ Then calculate the solution where $x + 10$ is positive: $ x + 10 = \dfrac{1}{3} $ Subtract ${10}$ from both sides: $ \begin{eqnarray} x + 10 &=& \dfrac{1}{3} \\ \\ {- 10} && {- 10} \\ \\ x &=& \dfrac{1}{3} - 10 \end{eqnarray} $ Change the ${ - 10}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{1}{3} {- \dfrac{30}{3}} $ $ x = -\dfrac{29}{3} $ Thus, the correct answer is $x = -\dfrac{31}{3} $ or $x = -\dfrac{29}{3} $.